29.8 Example: Endangered species

A study of endangered species (Harnish and Nataraajan 2020) examined

…whether perceived physical attractiveness of a species impacted participants’ attitudes toward supporting and protecting the species…

Harnish and Nataraajan (2020), p. 1703

To do so, 210 undergraduate students were surveyed about 14 animals on various aspects of supporting and protecting them. Part of the data are summarised below, for two animals when asked about ‘support to protect the animal from illicit trade.’ Larger values means greater support for protecting the animal from illicit trade.

Species Mean score Standard deviation
Bay Checkerspot Butterfly 3.10 1.06
Valley Elderberry Longhorn Beetle 2.33 1.13
Difference 0.77 1.07

(Notice that the standard deviation of the difference is not the difference between the two given values of the standard deviation.)

The difference is defined as each student’s score for the butterfly (deemed more attractive) minus their score for the beetle (deemed less attractive). A positive value therefore means more support (on average) for the butterfly.

The RQ is whether there is a mean difference between support for each animal, so the parameter is \(\mu_d\), the population mean difference.

The researchers wished to test if

…animals perceived as more physically attractive (i.e., the butterfly) compared to those which are perceived as less physically attractive (i.e., the beetle) will receive relatively more support to prevent the species from illicit trade

Harnish and Nataraajan (2020), p. 1704

Given how the difference are defined, the hypotheses are:

  • \(H_0\): \(\mu_d = 0\)
  • \(H_1\): \(\mu_d > 0\) (i.e., one-tailed, based on the researchers’ purpose)

The mean difference is \(\bar{d} = 0.77\) and \(s_d = 1.07\). The value of \(\bar{d}\) will vary from sample to sample, so has a standard error:

\[ \text{s.e.}(\bar{d}) = \frac{s_d}{\sqrt{n}} = \frac{1.07}{\sqrt{210}} = 0.073837. \]

The value of the test statistic is

\[ t = \frac{\bar{d} - \mu_d}{\text{s.e.}(\bar{d})} = \frac{0.77 - 0}{0.073837} = 10.43, \] which is a very large value. Hence, the \(P\)-value will be very small, certainly less than \(0.05\).

Since the sample size is much larger than 25, the test will be statistically valid.

We write:

There is very strong evidence (\(t = 10.43\); one-tailed \(P<0.001\)) that the mean difference in support for protecting the Bay Checkerspot Butterfly from illicit trade is greater than support for protecting the Valley Elderberry Longhorn Beetle from illict trade (mean difference: 0.77; standard deviation: 1.07; 95% CI for the difference: 0.62 to 0.92).


Harnish RJ, Nataraajan R. Attitudes toward wildlife: The impact of physical attractiveness. Psychology & Marketing. Wiley Online Library; 2020;37(12):1703–7.